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Home / Differential Geometry / Curve in space|Differential Geometry

Curve in space|Differential Geometry

Space curve: A space curve is defined as the locus of a point whose cartesian coordinates are the functions of a single variable parameter u, say.

When the curve is not a plane curve, it is said to skew, twisted or tortuous.

Equation of space curve: The equation of the space curve at the point u is given by

x = x(u) ————————(i)

where s is any vector of any point on the curve corresponding to the parameter u, say the point P on the curve x = x(u) is generally denoted by P(u) corresponding to the value of parameter u.

Also the equation of the space curve at the point s is given by

x = x(s) ———————–(ii)

where x is any vector of any point on the curve corresponding to the arc length s (say).

Example: The following are the equation of space curve

(i) x = x(u)

(ii) x = (cos u, sin u, cu)

Tangent line to a curve: The tangent line to a curve at a point P(u) is defined as the limiting position of a straight line L through P(u) and neighbouring point Q(u +du) on the curve as Q approaches P (i.e., Q → P) along the curve.

c1

Unit tangent vector to a space curve:

Let P ≡(x, y, z),

Q≡(x +dx, y+dy, z+dz),

= s, = ds, and

= s + ds .

Let P and Q be two neighbouring points on the curve C whose position vectors are x and x + dx referred to the origin O. Now from OPQ, we have

c2

c3

= x+dx-x

= dx

Let A be a fixed point on the space curve and arc AP = s and are AQ = s + ds, so that PQ = ds. Now, unit vector along chord PQ is

c4

Now if the point Q tends to P, then the chord PQ tends to be a tangent at P. Thus we have

c5

Hence unit vector along tangent at P

c6

Hence if the unit tangent vector at P is t, then we have

c7

Osculating plane (or plane of curvature): Let Q, P, R be three consecutive points on a space curve. The plane passing thorough Q, P, R when Q → P and R→P is called the osculating plane at P on the space curve.

c8

Equation of osculating plane: Let the equation of the space curve be x = x(u) and Q(u2), P(u1) and R(u3) be three consecutive points on this space curve. Let the equation of the plane passing through Q(u2), P(u1) ,R(u3) be

y(a).a = p,

where a is a constant vector.

So, we have

y(u1).a = p, y(a2).a =p, y(u3).a = p

⟹ y(u1).a –p=0, y(a2).a –p=0, y(u3).a –p=0

Also x= x(u) curve lies on thise plane, so we have

x(u1).a –p=0, x(u2).a –p=0, x(u3).a –p=0

∴ x(u).a = 0 at u1, u2, u3

Suppose that f(u) = x(u).a-p=0

Thus at at u1, u2, u3, we have

f(u1)=0, f(u2)=0, f(u3)=0.

Hence according to Rolle’s theorem, we have

f(v1)=0, u1<v1<u2

f(v2)=0, u1<v2<u2.

Now f(v) is continuous in [v1, v2] and hence again applying Rolle’s, we get

f(v3)=0, v1<v3<v2.

When the points Q(u2), R(u3) i.e., u2, u3 approach u1, then all the points u2, u3, v1,v2, v3 →u1

Now the equation of the plane is

y(a). a – p=0 ——————(i)

Also,y(u1) = x(u1).a –p=0 ——————–(ii)

ẏ (u) = ẋ(u1).a =0 —————————(iii)

ӱ (u) = ẍ(u1).a =0 ————————–(iv)

Subtracting (ii) from (i), we get

y(u).a- x(u1).a =0

⟹{ y(u)- x(u1)}.a =0 ———————-(v)

Replacing u1 by u in (v) , (iii) and (iv), we get

y(u)- x(u)}.a =0, ẋ(u).a =0, ẍ(u).a =0

⟹ (y-x) = 0, ẋ . a =0, ẍ . a =0   where x = x(u)

Since all the vectors (y-x), ẋ, ẍ are perpendicular to the constant vector a, so (y-x), ẋ, ẍ lies on the same plane.

Hence [(y-x) ẋ ẍ] = 0——————(vi)

which is the equation of osculating plane.

If y = (X, Y, Z) and x = (x, y, z), then substituting these values in (vi), the equation of the osculating plane given by

c10

 

 

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